Csaba Vincze (2013)
University of Debrecen
A recent trend in convex geometry is to investigate intersections of sets under a weaker condition than convexity. In 2001 N. A. Bobylev  provided a star-shaped set analogue of Helly's theorem. Especially Bobylev proved that if we have a family B of compact sets in the coordinate space of dimension n and every n+1 (not necessarily different) members of B have a star-shaped intersection then the intersection of all elements in B is star-shaped. The star- shaped set analogue of Klee's theorem 4.1.6 is true as well. In fact the result due to Marilyn Breen  states that the associated intersection is non-empty, star-shaped and its kernel is at least k-dimensional for an appropriate choice of k between 0 and n.
Theorem 6.1.1 Consider a collection B of sets in the coordinate space of dimension n and let
be a fixed integer. If every countable subfamily containing not necessarily different members of B has a star-shaped intersection whose kernel is at least k-dimensional then the intersection of all the members in B is a star-shaped set whose kernel is at least k-dimensional.
The proof is actually an induction on the dimension of the embedding space. In case of the coordinate line of dimension n=1 any star-shaped set isconvex and, by Klee's theorem 4.1.6, the intersection of all the members in B is non-empty and convex (especially star-shaped). If the intersection contains at least two different points then it is of dimension 1 and the proof is finished independently of the values k=0 or 1. Suppose now that the intersection is a singleton:
(for the sake of simplicity countable subfamilies will be labelled by natural numbers as usual to avoid double superscripts). We are going to prove that there exists a sequence of the members in B whose intersection reduces to a single point too. According to 6.1 for any natural number m there exists a set B(m) in B such that
In a similar way let B(- m) be a set in B such that
By our assumption the intersection
is star-shaped (especially convex) which means that it must be reduced to a singleton. Otherwise we have a segment in 6.2 containing x together with points having arbitrarily small rational distance 1/m from x which is a contradiction. Suppose that the statement is true for the coordinate space of dimension at most n - 1. By the help of the inductive hypothesis one can prove the following lemma which is the key step in the proof of theorem 1.1.
Lemma 6.1.2 Under conditions of theorem 6.1.1 the intersection of all the members in B is non-empty and it contains a k-dimensional convex subset.
Using 6.1.2 the proof of theorem 6.1.1 can be finished as follows. Let
be the intersection of all the members in B. We should check that V is star-shaped and its kernel is at least k-dimensional. Consider a new collection
M(γ) contains the points of B(γ) from which any point in V can be visible. Consider a countable subfamily M(1), ..., M(m), ... together with the associated sets B(1), ..., B(m), ... and let
be the intersections of the corresponding countable family of sets. Choose a point p in Ker K. Since V is a subset in K
for any element v in V and i=1, 2, ... Therefore p is in M. Especially M is a subset in K. This means that any element of M is visible from p, i.e.
Relation 6.4 says that we can apply lemma 6.1.2 to the collection 6.3:
and the intersection contains an at least k-dimensional convex subset. To finish the proof observe that
because the left hand side contains just the points of V from which any point in V can be visible, cf. 6.3. Therefore V is a star-shaped set and Ker V contains an at least k-dimensional convex subset. The proof of the key lemma 6.1.2 is based on the inductive hypothesis. Two different cases should be considered.
I. First case. Suppose that for some countable subfamily B(1), ..., B(m), ... the intersection K of the sets is at most of dimension n - 1. Then the maximal dimension of a convex subset M in K is also less than the dimension of the embedding space. Let B(1), ..., B(m), ... be chosen in such a way thatthe following minimax condition is satisfied: the maximum of the dimension of convex subsets in the intersection is as small as possible. This means that if we can inscribe a τ-dimensional convex subset into K as that of maximal dimension, then the intersection of any countable subfamily contains an at least τ-dimensional convex subset.
Lemma 6.1.3 If the family
satisfies the minimax condition then
where M is a convex subset of maximal dimension in K.
Proof Suppose, in contrary, that we have a point p in Ker K which is not in the affine hull of M. Then the convex hull of the union of M and p would be a greater dimensional convex subset in K than M.
Conditions for M are
(M1) M is convex,
(M2) M is a subset in K,
(M3) M is of dimension τ.
Consider the intersection of the affine hulls of M as it runs through the subsets satisfying M1, M2 and M3. We have by lemma 6.1.3 that
where H is an affine subspace of dimension at most τ. Suppose that the dimension of the affine subspace H associated with the countable family B(1), ..., B(m), ... satisfying the minimax condition is as great as possible. Choose an arbitrary countable subfamily B*(1), ..., B*(m), ... with intersection K* and let
be the union of the subfamilies with intersection
Recall that T contains a convex subset of dimension at least τ but T is a subset in K. Therefore 6.6 also satisfies the minimax condition. Thus
for the dimension of the associated affine subspace
where M' is a τ-dimensional convex subset in T. Especially, M' is a subset in K. Therefore
showing that H=L, i.e. the associated affine subspaces coincide because of the maximality condition for the dimension of H. On the other hand lemma 1.3 implies again that Ker T is a subset in L and, consequently,
Finally we apply the inductive hypothesis to the family of sets
in the coordinate space of dimension at most τ. The intersection of sets in 6.7 is contained in the intersection of all the members in B together with its k-dimensional kernel which is a convex set.
II. Second case. Suppose that for every countable subfamily has an n-dimensional intersection and let
be the collection of intersections of countable subfamilies. It can be easily seen that
and for any countable subfamily
we have a corresponding countable subfamily
for any index j. The intersection of 6.9 is just that of 6.10. Therefore the intersection of 6.9 has a non-empty interior and the same is true for the intersection of the closures of sets in 6.9. Let T be the intersection of the closures of sets in P (cf. theorem 4.1.5). Since the complement of T is expressed as the union of open subsets we can choose, by Lindelöf's theorem, a countable subcover. Taking the complement again T can be expressed as the intersection of a countable subfamily (say 6.9) of the closures:
Since the interior of T is non-empty it contains an open ball D in its interior. If D is a subset of the intersection of sets without closure operator the proof is finished. Otherwise suppose that p is a point in D such that p is not in G(1) (but p is in the closure of G(1)). Since G(1) is represented as a countable intersection of sets from B we can suppose thatp is not in B(1) for some set in B. Consider a countable subfamily B(1), ..., B(m), ... of subsets in B. The intersection
is a star-shaped set. Let z(1) be a point of Ker K. Since p is not in B(1) we have that p is not in K and the ray emanating from p into the opposite direction relative to z(1) does not contain points from K because they would be visible from z(1) together with p.
Recall that p is an interior point of the intersection of the closures of sets in P and thus it is an interior point of the closure of K (belonging to P). So we have a segment S(1) on the line of p and z(1) such that it has no common points with K but it is in the interior of the closure of K. If the kernel of K is a subset of aff S(1) we are ready. Otherwise we can construct a two-dimensional triangle S(2) such that it is in the interior of the closure of K but there are no common points with K.
If the kernel of K is a subset of aff S(2) we are ready. Otherwise repeat the process to construct a tetrahedron S(3) such that it has no common points with K but it is in the interior of the closure of K.
Repeat the algorithm as far as possible we can construct a j-dimensional simplex S(j) such that it has no common points with K but it is in the interior of the closure of K. By the constructing process,
Let τ be the number of the maximal dimension of sets in D which are disjoint from K. Then τ is greater than or equal to k but it must be less than n. It is clear because in case of τ=n there would be an n-dimensional simplex S which is disjoint from K (i.e. the interior points of S could not be limits of sequences in K) but S is a subset ofthe closure of K. Therefore
Suppose that the collection B(1), ..., B(m), ... has the greatest possible value of τ and consider the family of sets
where H is the affine hull of S(τ). Using the inductive hypothesis, the intersection
is a star-shaped set with kernel of dimension at least k and the proof is finished.